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Question

Phân tích đa thức: 7x3 - 7x - 6x3 - 6      thành nhân tử

Trần Thanh Phương · Accepted Answer

$7x^3-7x-6x^3-6$

$=7x\left(x^2-1\right)-6\left(x^3+1\right)$

$=7x\left(x-1\right)\left(x+1\right)-6\left(x+1\right)\left(x^2-x+1\right)$

$=\left(x+1\right)\left[7x\left(x-1\right)-6\left(x^2-x+1\right)\right]$

$=\left(x+1\right)\left(7x^2-7x-6x^2+6x-6\right)$

$=\left(x+1\right)\left(x^2-x-6\right)$

$=\left(x+1\right)\left(x^2-3x+2x-6\right)$

$=\left(x+1\right)\left[x\left(x-3\right)+2\left(x-3\right)\right]$

$=\left(x+1\right)\left(x-3\right)\left(x+2\right)$Câu trả lời: $7x^3-7x-6x^3-6$

$=7x\left(x^2-1\right)-6\left(x^3+1\right)$

$=7x\left(x-1\right)\left(x+1\right)-6\left(x+1\right)\left(x^2-x+1\right)$

$=\left(x+1\right)\left[7x\left(x-1\right)-6\left(x^2-x+1\right)\right]$

$=\left(x+1\right)\left(7x^2-7x-6x^2+6x-6\right)$

$=\left(x+1\right)\left(x^2-x-6\right)$

$=\left(x+1\right)\left(x^2-3x+2x-6\right)$

$=\left(x+1\right)\left[x\left(x-3\right)+2\left(x-3\right)\right]$

$=\left(x+1\right)\left(x-3\right)\left(x+2\right)$

Hoaa · Answer

$7x^3-7x-6x^3-6$

=$\left(7x^3-7x\right)+\left(-6x^3-6\right)$

=$7x\left(x^2-1\right)-6\left(x^3+1\right)$

=$7x\left(x-1\right)\left(x+1\right)-6\left(x+1\right)\left(x^2-x+1\right)$

=(x+1)(x-1)(7x-6)(x^2-x+1)

=(x^2-1)(7x-6)(x^2-x+1)Câu trả lời: $7x^3-7x-6x^3-6$

=$\left(7x^3-7x\right)+\left(-6x^3-6\right)$

=$7x\left(x^2-1\right)-6\left(x^3+1\right)$

=$7x\left(x-1\right)\left(x+1\right)-6\left(x+1\right)\left(x^2-x+1\right)$

=(x+1)(x-1)(7x-6)(x^2-x+1)

=(x^2-1)(7x-6)(x^2-x+1)

Bài 9: Phân tích đa thức thành nhân tử bằng cách phối hợp nhiều phương pháp